# lenghts of sides triangle

• Dec 11th 2009, 08:37 AM
wolfhound
lenghts of sides triangle
Hello,
Can you tell me how to find the sides of he triangles in Q1 and Q2 in the attached drawing
they are right angled triangles
in question 2 i found that a=1.9 and b=1.3 using Soh

but i want to know If i can use $\displaystyle a^2 = b^2 + c^2 - 2bc(CosA)$

I want to be able to use other forumlas than a^2 + b^2=c^2

• Dec 11th 2009, 09:40 AM
Stuck Man
I don't think any values can be found for example 2.

The first example has a right angle and you don't want to use more complicated methods than is necessary so the Pythagoras theorem will do or the basic trigonometrical ratios.
• Dec 11th 2009, 11:28 AM
wolfhound
There is answers! I dont know how to get them someone help!!!!!!!! (Headbang)
• Dec 11th 2009, 01:53 PM
skeeter
Quote:

Originally Posted by wolfhound
Hello,
Can you tell me how to find the sides of he triangles in Q1 and Q2 in the attached drawing
they are right angled triangles
in question 2 i found that a=1.9 and b=1.3 using Soh

but i want to know If i can use $\displaystyle a^2 = b^2 + c^2 - 2bc(CosA)$

I want to be able to use other forumlas than a^2 + b^2=c^2

as stated previously by the Stuck Man ... question 2 does not have enough information.

you need one more side or one more angle. there are an infinite number of right triangles with hypotenuse length 2.3
• Dec 12th 2009, 02:22 AM
wolfhound
Sorry,
The angle B in question 2 is 34 degrees
Please tell me how to calculate the answers for the two questions,I have an important exam next week, can it be done with the other formula in my original post?I can seem to do them with that?is that formula only for triangles that are not 90*
• Dec 12th 2009, 02:28 AM
mr fantastic
Quote:

Originally Posted by wolfhound
Sorry,
The angle B in question 2 is 34 degrees
Please tell me how to calculate the answers for the two questions,I have an important exam next week, can it be done with the other formula in my original post?I can seem to do them with that?is that formula only for triangles that are not 90*

Q1. It's an isosceles triangle therefore c = 1.

Q2. You should know this:

$\displaystyle \sin 34^0 = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{b}{2.3} \Rightarrow ....$

$\displaystyle \cos 34^0 = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{a}{2.3} \Rightarrow ....$
• Dec 12th 2009, 04:36 AM
wolfhound
Quote:

Originally Posted by mr fantastic
Q1. It's an isosceles triangle therefore c = 1.

Q2. You should know this:

$\displaystyle \sin 34^0 = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{b}{2.3} \Rightarrow ....$

$\displaystyle \cos 34^0 = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{a}{2.3} \Rightarrow ....$

Yes thats how I got the sides on question 2.

But I want to know how to work out the other side is 1(in question 1). In the exam I cant just write 1,I need to prove it

When is this formula used $\displaystyle a^2 = b^2 + c^2 - 2bc(CosA)$
and why cant I find the other side using this?
• Dec 12th 2009, 04:52 AM
mr fantastic
Quote:

Originally Posted by wolfhound
Yes thats how I got the sides on question 2.

But I want to know how to work out the other side is 1(in question 1). In the exam I cant just write 1,I need to prove it

When is this formula used $\displaystyle a^2 = b^2 + c^2 - 2bc(CosA)$
and why cant I find the other side using this?

Did I not I say
Quote:

It's an isosceles triangle
which, I would have thought, provides the necessary justification.

Alternatively, use Pythagoras' Theorem. Alternatively, use $\displaystyle \sin 45^0 = \frac{\text{opposite side}}{\text{hypotenuse}}$ (which you should have realised from what I posted about question 2).

Why are you insisting on wanting to use the cosine rule? For a right angle triangle A = 90 degrees and it's just Pythagoras' Theorem. The cosne rule and the sine rule are used when you have a triangle that is NOT a right-triangle. You should know that.
• Dec 12th 2009, 06:55 AM
wolfhound
Quote:

Originally Posted by mr fantastic
Did I not I say

which, I would have thought, provides the necessary justification.

Alternatively, use Pythagoras' Theorem. Alternatively, use $\displaystyle \sin 45^0 = \frac{\text{opposite side}}{\text{hypotenuse}}$ (which you should have realised from what I posted about question 2).

Why are you insisting on wanting to use the cosine rule? For a right angle triangle A = 90 degrees and it's just Pythagoras' Theorem. The cosne rule and the sine rule are used when you have a triangle that is NOT a right-triangle. You should know that.

I am curious why I could not get a correct answer using the cosine rule? I thought it would work on right angled triangles also? NO?
I want to get in practice with the different formulas that is all (Clapping)
• Dec 13th 2009, 07:52 PM
Greek2Me
Here are links to tutorials for basic "right angle" trig, with a very DIFFERENT approach than the usuals "soh cah toa" mneumonic. The first shows you how to make a "Trig Tool", which elimininates memorizing a lot of formulas, the next two show you how to USE it.

How to Memorize the Trig Functions (Without Losing Your Mind!) | eHow.com
How to Pass That Trigonometry Exam (Without Losing Your MIND!) Part I - Unknown Sides | eHow.com
How to Pass That Trigonometry Exam (Without Losing Your MIND!) Part II - Unknown Angles | eHow.com